UVA 10828 Back to Kernighan-Ritchie(高斯消元)

高斯消元求概率

对于非起点，期望x[i] = ∑x[j] / deg[j]

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<utility>
using namespace std;
typedef long long LL;
const int N = 108, INF = 0x3F3F3F3F;
const double eps = 1e-8;
double f[N][N];
std::vector<int> g[N];
bool inf[N];

template<typename T>
void gauss(T A[N][N], int n){
    for(int i = 0; i < n; i++){
        int r = i;
        for(int j = i + 1; j < n; j++){
            if(abs(A[j][i]) > abs(A[r][i])){
                r = j;
            }
        }
        if(abs(A[r][i]) < eps){
            continue;
        }
        if(r != i){
            for(int j = 0; j <= n; j++){
                swap(A[r][j], A[i][j]);
            }
        }
        for(int k = 0; k < n; k++){
            if(k != i){
                for(int j = n; j >= i; j--){
                    A[k][j] -=  A[k][i] / A[i][i] * A[i][j];
                }
            }
        }
    }
}

int main(){
    int cas = 0, n, q;
    while(~scanf("%d", &n) && n){
        for(int i = 0; i <= n; i++){
            g[i].clear();
        }
        int u, v;
        while(scanf("%d %d", &u, &v) &&  u && v){
            u--;
            v--;
            g[u].push_back(v);
        }
        memset(f, 0, sizeof(f));
        for(int i = 0; i < n; i++){
            f[i][i] = -1;
            for(int j = 0; j < g[i].size(); j++){
                f[g[i][j]][i] += 1.0 / (double)g[i].size();
            }
        }
        f[0][n] = -1;
        gauss(f, n);
        memset(inf, 0, sizeof(inf));
        for(int i = n - 1; i >= 0; i--){
            if(abs(f[i][i]) < eps && abs(f[i][n]) > eps){
                inf[i] = 1;
            }
            for(int j = i + 1; j < n; j++){
                if(inf[j] && abs(f[i][j]) > eps){
                    inf[i] = 1;
                    break;
                }
            }
        }
        printf("Case #%d:
", ++cas);
        scanf("%d", &q);
        while(q--){
            int u;
            scanf("%d", &u);
            u--;
            if(inf[u]){
                printf("infinity
");
            }else{
                printf("%.3f
", (abs(f[u][u]) < eps )? 0.0 : abs(f[u][n] / f[u][u]));
            }
        }

    }
    return 0;
}